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Prove that $$|z_1 + z_2| + |z_2 + z_3| + |z_3 + z_1| \le |z_1 + z_2 + z_3| + |z_1| + |z_2| + |z_3|$$

$$\begin{align} &\ \ |z_1 + z_2 + z_3| + |z_1| + |z_2| + |z_3| \\ =&\ \ |(z_1 + z_3) + (z_2 + z_3) - z_3| + |z_1| + |z_2| + |z_3| \\\ge &\ \ |(z_1 + z_3) + (z_2 + z_3)| - |z_3| + |z_1| + |z_2| + |z_3| \\=&\ \ |(z_1 + z_3) + (z_2 + z_3)| + |z_1| + |z_2| \\\ge& \ \ |(z_1 + z_3) + (z_2 + z_3)| + |z_1 + z_2| \end{align}$$

  • I am stuck here. Please help me complete the proof.
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    Why not search on the web for the proof? For instance, a good proof is found in here http://www.cut-the-knot.org/arithmetic/algebra/Hlawka.shtml – 2017-02-17
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    @ΘΣΦGenSan Because I want to know how can I continue from here. – 2017-02-17
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    I would advise that you not disregard a good proof because you feel like you have to continue along the line of thinking that you've already started on. It would almost certainly take you less time to learn to apply the techniques involved in the pre-existing proof (and hopefully to recognize when to apply them next time). – 2017-02-17
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    @MichaelLee Since when I disregard a good proof. I just thought that we can easily write a proof from where I left. I see nothing wrong in it. – 2017-02-17

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We need to prove that $$\left(|a+b|+|a+c|+|b+c|\right)^2\leq\left(|a+b+c|+|a|+|b|+|c|\right)^2$$ and since $$|a+b|^2+|a+c|^2+|b+c|^2=|a+b+c|^2+|a|^2+|b|^2+|c|^2,$$ it's enough to prove that $$\sum_{cyc}\left(|a(a+b+c)|+|bc|\right)\geq\sum_{cyc}|(a+b)((a+c)|,$$ which is true because $$|a(a+b+c)|+|bc|\geq|a(a+b+c)+bc|=|(a+b)(a+c)|$$ Done!

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    Why is $|a+b|^2+|a+c|^2+|b+c|^2=|a+b+c|^2+|a|^2+|b|^2+|c|^2$ true for complex numbers? – 2017-09-05
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    @Taha Akbari Let $a=a_1+a_2i$, $b=b_1+b_2i$ and $c=c_1+c_2i$, where $\{a_1,a_2,b_1,b_2,c_1,c_2\}\subset\mathbb R$. Thus, easy to see that $(a_1+b_1)^2+(a_2+b_2)^2+(a_1+c_1)^2+(a_2+c_2)^2+(b_1+c_1)^2+(b_2+c_2)^2=a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2+(a_1+b_1+c_1)^2+(a_2+b_2+c_2)^2$. – 2017-09-05